(chapter 6) Example: A gas stream consisting of nhexane in methane is fed to a condenser at 60°C and 1.2 atm. The dew point of the gas (considering hexane as the only condensable component) is 55°C. The gas is cooled to 5°C in the condenser, recovering pure hexane as a liquid. The effluent gas leaves the condenser saturate with hexane at 5°C and 1.1 atm and is fed to a noiler furnace at a rate of 207.4 L/s. a) Calculate the mole fracions of hexane in the condenser feed and product gas streanms and the rate of hexane condensation.Initial sketch: Flow diagram: Our Table has the form:
Now, we fill in the table. The ideal gas law in stream two, with 1.1 atm, 207 L/s, and 278.25 K gives F2 total.
Now, additionally, if we consider Raoult's Law for stream 2, a vapor mixture, which is in equilibrium with stream 3, pure nHexane liquid. Raoult's Law is, again: x_{3H} = 1, P = 1.1 atm P_{H}* = something we know, we are given the temperature, so we can calculate this with Antoines Equation, and from the appendix B4, we have. P_{H}* = 10^(a  b/(c + T(in °C)), with a = 6.88555, b = 1175.817, c = 224.867, and T = 5, this gives and answer in mmHg (58.93). Thus now have y_{2H} from Raoults law (.07049), and also, since y_{2M} = 1  y_{2H}, we have y_{2H} = .9295, and thus,
and, furthermore,
Now, we can apply Raoult's Law to the inlet stream. This seems weird, cause there is only vapor and it is not at it's dew point, but because we know the dew point of the vapor and because there is only one condensible species, we can recall Now, as before, we have P (=1.2 atm), and P* (= 483.3 mmHg, calculated from Antoines equation, a,b, and c given above, and T = 55’C), so we can calculate y_{H1} = .5299, and thus y_{M1} = .4701. And it follows that,
furthermore,

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